I'm always trying to find the integral representation of $\pi$ using some interesting special function, at this time I have got the below representation $$I=\int_{-2}^{2} \tan^{-1} \bigg( \exp(-x^2\text{erf}(x)) \bigg) \;dx=\pi$$ and according to Wolfram alpha its numerical value is very close to $\pi$.

The problem that I have accrossed is the closed form of : $\exp(-x^2\text{erf}(x))$ in the range $[-2,2]$ , I have used $$\mathrm{erf}\!\left(x\right)^2\approx1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha x^2}{1+\beta x^2}\,x^2 \Big)$$ $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi }$$ $$\beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }.$$ The value of the corresponding error function is $1.1568\times 10^{-7}$ that is to say almost $250$ times smaller than with the initial formulation. The maximum error is $0.00035$. But when we try to replace that approximation in $\tan^{-1}$ for evaluation the formula would be complicat to integrate it. My question here is if this result is known. Is it $\pi$? If no any simple way for integration ?

  • $\begingroup$ That approximation of the erf is out by a factor of 250!? That's not a very good approximation then ... or have I misunderstood you? $\endgroup$ – AmbretteOrrisey Dec 3 '18 at 22:03

Yes, the integral $I$ is equal to $\pi$. Note that for $t>0$ $$\arctan(t)+\arctan(1/t)=\pi/2.$$ and after letting $y=-x$ we get $$I:=\int_{-2}^{2} \tan^{-1} \bigg( \exp(-x^2\text{erf}(x)) \bigg) \;dx =\int_{-2}^{2} \tan^{-1} \bigg( \exp(y^2\text{erf}(y)) \bigg) \;dy.$$ Hence $$I=\frac{1}{2}\int_{-2}^{2}\arctan(t(x))+\arctan(1/t(x))dx=\frac{\pi/2\cdot 4}{2}=\pi$$ where $t(x)=\exp(-x^2\text{erf}(x))$.

The same argument holds if we replace $-x^2\text{erf}(x)$ with any odd function (see Michael Seifert's comment below).

  • 1
    $\begingroup$ Note that this argument would still work, and the integral would still be $\pi$, if you replaced $x^2 \mathrm{erf}(x)$ by any odd function $f(x)$. $\endgroup$ – Michael Seifert Dec 3 '18 at 20:01
  • $\begingroup$ Attractive answer $\endgroup$ – zeraoulia rafik Dec 3 '18 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.